A particle is confined in a potential well such that its allowed energies are $E^n = n^2\epsilon$, where $n = 1, 2, \dots$ is an integer and $\epsilon$ a positive constant. The corresponding energy eigenstates are $\lvert1\rangle, \lvert2\rangle, \dots , \lvert n\rangle, \dots$ At t = 0 the particle is in the state:

$\lvert\psi(0)\rangle = 0.2\lvert1\rangle + 0.3\lvert2\rangle + 0.4\lvert3\rangle + 0.843\lvert4\rangle$.

What is the probability if energy is measured at $t=0$ of finding a number smaller than $6\sigma$?

Am I right in saying this would just be the sum of states $n = 1$ and $n = 2$ which is $0.5$?

Then I'm wondering how you would calculate the mean value and rms deviation of the energy of the particle in the state $\lvert\psi(0)\rangle$

How do I find the state vector $\lvert\psi\rangle$ at any time $t$? And therefore do the results calculated above remain valid for any arbitrary time?

The last thing I'm stuck on is, lets say the energy is measured and it is said to be $16\epsilon$. After this measurement, what is the state of the system, and what result would you get if you tried to measure energy again?


1 Answer 1

  1. For your first question-you actually have to square the coefficients and add them up. Notice that it is the sum of the squares of the coefficients which add up to unity. So the required probability would be $(0.2)^2+(0.3)^2=0.13$.

  2. In general, energy superpositions are not stationary states, so your state-vector will change with time. However, the probabilities will remain unchanged, because when you evolve the state-vector with time, each term in its expansion will only vary by a phase factor of the form $e^{-ikt}$, where t is time and k is some constant. You can see that the modulus will remain the same, since $|ce^{-ikt}|^2=|c|^2.$

  3. The mean value of energy will just be $\frac{\sum|c_n|^2E_n}{4}$.

  4. To answer your last question: Energy eigenstates are stationary states. Therefore, it will remain in $E^n$ after you measure it, and you will get the same result if you measured energy again.

  • $\begingroup$ Thank you so much this is super helpful! So how would I actually go about calculating the state vector at any time t? Is it just $|ψ(t)> = e^{−iφ(t)} |ψ(0)>$? $\endgroup$ Commented Dec 2, 2020 at 16:22
  • $\begingroup$ We usually separate the space and the time parts of the wavefunction, so one would write $\psi(x,t)=\phi(x)f(t)$, where $f(t)$ will just be the phase factor evolving it in time. By the way, do mark my answer as correct if you found that it sufficed. $\endgroup$
    – A.D.
    Commented Dec 2, 2020 at 16:29

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